Problem 9 - Applied Discrete Structures

Question

(a) Let \(A, B,\) and \(D\) be \(n \times n\) matrices. Assume that \(B\) is invertible. If \(A=B D B^{-1},\) prove by induction that \(A^{m}=B D^{m} B^{-1}\) is true for \(m \geq 1\) (b) Given that \(A=\left(\begin{array}{ll}-8 & 15 \\ -6 & 11\end{array}\right)=B\left(\begin{array}{ll}1 & 0 \\ 0 & 2\end{array}\right) B^{-1}\) where \(B=\) \(\left(\begin{array}{ll}5 & 3 \\ 3 & 2\end{array}\right)\) what is \(A^{10} ?\)

Step-by-Step Solution

Answer

The matrix \(A^{10}\) is \(\begin{pmatrix} -10067 & 20460 \\ -8136 & 16593 \end{pmatrix}\).

Step 1: Base Case for Induction

To prove by induction, we first verify the base case of \(m = 1\). For \(m = 1\), we have \(A = BD B^{-1}\), which is true by the assumption given in the problem.

Step 2: Inductive Hypothesis

Assume that the statement is true for some integer \(m = k\), that is \(A^k = B D^k B^{-1}\). This is the inductive hypothesis.

Step 3: Inductive Step

We need to prove that \(A^{k+1} = B D^{k+1} B^{-1}\). Starting from the inductive hypothesis, \(A^{k+1} = A^k A = (B D^k B^{-1})(B D B^{-1})\). Using the associativity of matrix multiplication, this becomes \(B D^k (B^{-1} B) D B^{-1} = B D^{k+1} B^{-1}\), since \(B^{-1} B = I\).

Step 4: Conclusion for Induction

Since both the base case and the inductive step hold, by induction, we have proven that for all \(m \geq 1\), \(A^m = B D^m B^{-1}\).

Step 5: Matrix Powers of D

Given the matrix \(D = \begin{pmatrix} 1 & 0 \ 0 & 2 \end{pmatrix}\), to find \(D^{10}\), we calculate the powers of the diagonal matrix as \(D^{10} = \begin{pmatrix} 1^{10} & 0 \ 0 & 2^{10} \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1024 \end{pmatrix}\).

Step 6: Compute A^{10}

Using the formula \(A^{10} = B D^{10} B^{-1}\), calculate \(A^{10} = \begin{pmatrix} 5 & 3 \ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 1024 \end{pmatrix} \begin{pmatrix} 5 & 3 \ 3 & 2 \end{pmatrix}^{-1}\).

Step 7: Find B^{-1}

Calculate the inverse \(B^{-1}\). The inverse of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\) is \(\frac{1}{ad-bc}\begin{pmatrix} d & -b \ -c & a \end{pmatrix}\). Here, \(ad-bc = 1\), so \(B^{-1} = \begin{pmatrix} 2 & -3 \ -3 & 5 \end{pmatrix}\).

Step 8: Final Calculation

Multiply the matrices: \(A^{10} = \begin{pmatrix} 5 & 3 \ 3 & 2 \end{pmatrix} \begin{pmatrix} 1 & 0 \ 0 & 1024 \end{pmatrix} \begin{pmatrix} 2 & -3 \ -3 & 5 \end{pmatrix}\). This results in \(A^{10} = \begin{pmatrix} -10067 & 20460 \ -8136 & 16593 \end{pmatrix}\).

Key Concepts

Matrix Powers

The concept of matrix powers is pivotal when dealing with continuous iterations or transformations represented by matrices. If you have a square matrix \( A \), the notation \( A^m \) indicates that the matrix is multiplied by itself \( m \) times. Each power of the matrix represents a further application of the transformation it describes.
For example:

• \( A^1 = A \)
• \( A^2 = A \cdot A \)
• \( A^3 = A \cdot A \cdot A \)

In the case of the problem, we considered an expression involving the power of another matrix \( D \) along with a transformation matrix \( B \). Understanding the behavior of diagonal matrices can often simplify complex calculations, as raising a diagonal matrix to a power only involves exponentiating the individual diagonal elements. This specificity often leads to easier computations, especially in cases where one component of the transformation remains unchanged.

Matrix Multiplication

Matrix multiplication is a fundamental operation in linear algebra, representing complex interactions across rows and columns of matrices. When you multiply two matrices, each element in the resulting matrix is derived from a dot product calculated between rows of the first matrix and columns of the second matrix. This process requires the number of columns in the first matrix to match the number of rows in the second matrix.

The result's dimension will match the number of rows of the first matrix by the number of columns of the second matrix. Matrix multiplication is associative and distributive, but notably, it is not commutative, meaning \( AB eq BA \) generally.

• Associative: \( (AB)C = A(BC) \)
• Distributive: \( A(B + C) = AB + AC \)
• Not Commutative: \( AB eq BA \)

In our case, the multiplication of matrices, including the inverse matrix and the power matrix, ensures the equivalence transformation necessary to progress further steps in calculations and proofs.

Inductive Proof

Inductive proofs are a powerful mathematical tool for demonstrating the validity of a statement across an infinite set of scenarios, often involving sequences or series. Induction involves two main parts:

• Base Case: Verify the statement holds for the initial case \( m = 1 \).
• Inductive Step: Assume it is true for seemingly arbitrary \( k \), and then prove it holds for \( k+1 \). This step transitions the claim from truth in one step to continuity for the subsequent step.

In our exercise, we've begun by affirming the base case (where \( m = 1 \) is validated by the given relation). The process then proceeds through the inductive step by logically progressing from \( A^k \) to \( A^{k+1} \) via matrix properties like associativity and identity. Once both are substantial, the principle of mathematical induction ensures truth for all subsequent \( m \) values greater than or equal to the base case.

Inverse Matrix Calculation

Calculating the inverse of a matrix is essential for solving matrix equations, finding unique solutions, and uncoupling transformations. For a matrix to be invertible, it must be a square matrix with a non-zero determinant.
The formula for finding the inverse of a 2x2 matrix \(\begin{pmatrix} a & b \ c & d \end{pmatrix}\), when the determinant \( ad-bc eq 0 \), is given by:\[\begin{pmatrix} d & -b \ -c & a \end{pmatrix} \frac{1}{ad-bc}\]
For our given matrix \( B = \begin{pmatrix} 5 & 3 \ 3 & 2 \end{pmatrix} \), the determinant is \( 1 \) making its inverse perfective. Knowing how to find this inverse empowered us to navigate through our calculations for \( A^{10} \), effectively solving for the transformation matrix's impact. This inverse serves as a mechanism to "untangle" the matrix multiplicative relationships seen in the problem.